Prove Z 1 0 arctan(x) x ln  1 + x2 (1 − x)2  dx = π3 16

Problem 12054

(American Mathematical Monthly, Vol.125, June-July 2018) Proposed by C. I. V˘alean (Romania).

Prove

Z 1 0

arctan(x)

x ln

 1 + x² (1 − x)²



dx = π³ 16.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. Let Hk=Pk j=1

1

j for k ≥ 1. We have that for x ∈ (0, 1), arctan(x) ln(1 + x²) = i

2(ln(1 − ix) − ln(1 + ix))(ln(1 − ix) + ln(1 + ix))

= i

2(ln²(1 − ix) − ln²(1 + ix)) = − Im ln²(1 − ix)
= −2 Im

∞

X

k=1

Hk(ix)^k+1 k + 1

!

where we used the fact that

−ln(1 − t) =

∞

X

k=1

t^k

k =⇒ −ln(1 − t) 1 − t =

∞

X

k=1

Hkt^k =⇒ ln²(1 − t) = 2

∞

X

k=1

Hkt^k+1 k + 1 . Hence

Z ¹

0

arctan(x) ln(1 + x²)

x dx = −2 Im

Z ¹

0

∞

X

k=1

Hki^k+1x^k k + 1 dx

!

= −2 Im

∞

X

k=1

Hki^k+1 k + 1

Z 1 0

x^kdx

!

= −2 Re

∞

X

k=1

Hki^k (k + 1)²

! . On the other hand,

Z 1 0

arctan(x) ln(1 − x)

x dx =

Z 1 0

∞

X

k=1

(−1)^kx^2kln(1 − x) 2k + 1 dx

=

∞

X

k=0

(−1)^k 2k + 1

Z ¹

0

x^2kln(1 − x) dx

=

∞

X

k=0

(−1)^k (2k + 1)²

Z ¹

0

ln(1 − x) d(x^2k+1−1)

= −

∞

X

k=0

(−1)^k (2k + 1)²

Z ¹

0

x^2k+1−1 x − 1 dx

= −

∞

X

k=1

(−1)^kH2k+1

(2k + 1)² = − Re

∞

X

k=0

Hk+1i^k (k + 1)²

! . Therefore

Z ¹

0

arctan(x)

x ln

 1 + x² (1 − x)²



dx = −2 Re

∞

X

k=1

Hki^k (k + 1)²

! + 2 Re

∞

X

k=0

Hk+1i^k (k + 1)²

!

= 2 Re

∞

X

k=0

i^k (k + 1)³

!

= 2 Im

∞

X

k=1

i^k k³

!

= 2 Im(Li3(i)) = π³ 16 because, by a known identity for the trilogarithm Li3,

2 Im(Li3(i)) = Im (Li3(z) − Li3(1/z))_z=i= −1

6Im ln³(−z) + π²ln(−z)

z=i= π³ 16.